Solve the following systems of homogeneous linear equations by matrix method:

3x – y + 2z = 0


4x + 3y + 3z = 0


5x + 7y + 4z = 0

The system can be written as



A X = 0


Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)


|A| = – 27 + 1 + 26


|A| = 0


Hence, the system has infinite solutions


Let z = k


3x – y = – 2k


4x + 3y = – 3k



A X = B


|A| = 9 + 4 = 13 ≠0 So, A – 1 exist


Now adj A = =


X = A – 1 B =


X =


Hence, x = , y = and z = k


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